My Math Forum  

Go Back   My Math Forum > High School Math Forum > Algebra

Algebra Pre-Algebra and Basic Algebra Math Forum


Reply
 
LinkBack Thread Tools Display Modes
November 6th, 2017, 10:37 AM   #1
Newbie
 
Joined: Nov 2017
From: Europe

Posts: 2
Thanks: 0

error: linear inequality (absolute function)

hello,

I tried to solve the following inequality:
abs(-2x+1)>2/3x+1

but get wrong results:

-2x+1>-2/3x-1

I add 1 then +2x then multiply with 3 and divide by 4 I get: x<3/2

and -2x+1>2/3x+1

cancel out 1 then +2x 0>8/3x so
I get x<0

surely this inequality is true for some x>0 - do you spot the error?


best,
richard

Last edited by skipjack; November 7th, 2017 at 12:30 AM.
dickgreenleaf is offline  
 
November 6th, 2017, 07:01 PM   #2
Global Moderator
 
greg1313's Avatar
 
Joined: Oct 2008
From: London, Ontario, Canada - The Forest City

Posts: 7,642
Thanks: 960

Math Focus: Elementary mathematics and beyond
There are two cases; one where -2x + 1 is positive and one where it's negative. It's zero at x = 1/2. Does that help? Hint: make a sign chart - if you know how. If not, consult Google (or a search engine of your choice).

You could also graph both functions on the same axes.

Last edited by greg1313; November 6th, 2017 at 07:10 PM.
greg1313 is offline  
November 7th, 2017, 05:07 AM   #3
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,824
Thanks: 752

Is the right side "(2/3)x+ 1" or "2/(3x)+ 1" or "2/(3x+ 1)"? Please use parentheses!
Country Boy is offline  
November 8th, 2017, 08:56 AM   #4
Newbie
 
Joined: Nov 2017
From: Europe

Posts: 2
Thanks: 0

Hi Country Boy, sorry that's the one I meant:
abs(-2x+1)>(2/3)x+1

Last edited by skipjack; November 8th, 2017 at 11:04 AM.
dickgreenleaf is offline  
November 8th, 2017, 11:20 AM   #5
Global Moderator
 
Joined: Dec 2006

Posts: 18,155
Thanks: 1422

Let's first consider x < 1/2, so that abs(-2x + 1) = -2x + 1.
To solve -2x + 1 > (2/3)x + 1,
add 2x - 1 to both sides to get 0 > (8/3)x,
which is equivalent to x < 0. So far, this matches what you did.

Now let's consider x $\small\geqslant$ 1/2, so that abs(-2x + 1) = 2x - 1.
To solve 2x - 1 > (2/3)x + 1,
subtract (2/3)x - 1 from both sides to get (4/3)x > 2,
which is equivalent to x > 3/2.
skipjack is offline  
November 10th, 2017, 04:09 AM   #6
Math Team
 
Joined: Jan 2015
From: Alabama

Posts: 2,824
Thanks: 752

By the way- this is NOT a "linear" inequality. Absolute value is not a linear function.
Country Boy is offline  
Reply

  My Math Forum > High School Math Forum > Algebra

Tags
absolute, error, function, inequality, linear



Thread Tools
Display Modes


Similar Threads
Thread Thread Starter Forum Replies Last Post
Maximum absolute error when calculating a difference using two rounded down values. trampolinerist Algebra 1 October 29th, 2017 07:16 PM
Triangle Inequality: Prove Absolute Value Inequality StillAlive Calculus 5 September 3rd, 2016 12:45 AM
Absolute Value Inequality with Absolute Values on Both Sides shiseonji Algebra 2 September 24th, 2013 09:36 AM
absolute value inequality ommmmid Algebra 5 March 19th, 2013 10:38 PM
least absolute error minimization with quadratic constraint benoit Linear Algebra 0 December 1st, 2010 03:34 AM





Copyright © 2017 My Math Forum. All rights reserved.